Read a string of alphanumeric characters after a ; - perl

Question

I'm teaching myself perl so I'm pretty new to this language. I've been reading over and over about regular expression but I can't figure out the right context. I want to do the following:
Let say I have a file name "testfile"
this files contains 3 lines,
test this is the first line
test: this is the first line
test; this is the third line
How can I read and print out only the third one and everything after the ; without the space. so basically "This is the third line"
This is what I'm thinking to do $string =~ m/this is the third/
This was edited incorrectly. In the first and second sentence there should a space before the test.in the third one shouldn't. So I want to skip the white space.

Answer 1

Grabbing from STDIN, it might look like this:
while ( <> ) {
print $1 if /^test; (.*\n)/;
}

Answer 2

You might find that YAPE::Regex::Explain is a handy tool:
Using Axeman's regular expression:
#!/usr/bin/env perl
use strict;
use warnings;
use YAPE::Regex::Explain;
my $expr = q(/^test; (.*\n)/);
print YAPE::Regex::Explain->new( $expr )->explain;
The regular expression:
(?-imsx:/^test; (.*\n)/)
matches as follows:
NODE EXPLANATION
----------------------------------------------------------------------
(?-imsx: group, but do not capture (case-sensitive)
(with ^ and $ matching normally) (with . not
matching \n) (matching whitespace and #
normally):
----------------------------------------------------------------------
/ '/'
----------------------------------------------------------------------
^ the beginning of the string
----------------------------------------------------------------------
test; 'test; '
----------------------------------------------------------------------
( group and capture to \1:
----------------------------------------------------------------------
.* any character except \n (0 or more times
(matching the most amount possible))
----------------------------------------------------------------------
\n '\n' (newline)
----------------------------------------------------------------------
) end of \1
----------------------------------------------------------------------
/ '/'
----------------------------------------------------------------------
) end of grouping
----------------------------------------------------------------------

Answer 3

If you only want the third line, then simply counting the lines and then doing:
s/.*;\s*//;
will remove everything until the ; and any white space after it. Note, however, that if the third line contains another ';' in it then you'll be in trouble. So if that's a possibility but there is no chance that one will exist earlier, then do this:
s/[^;]*;\s*//;
Which will delete only up until the first ';' (and trailing whitespace).
I suspect, however, in the long run you want to match all lines that contain some particular format and it won't always be "just the third". If that's the case then:
while(<>) {
if (/;\s*(.*)/) {
print $1;
}
}
Will get you closer to your end-goal.

Answer 4

Another way to achieve this is to try to remove everything up to the first ; (and any spaces straight afterwards), and only print the line if there was something to remove.
s/.*?;\s*//;
This line basically says: "match any characters (but as few as possible), then a semicolon, then any spaces, and replace it all with nothing".
You can then make a the program that reads from STDIN:
while (<>) {
print if s/.*?;\s*//;
}
You can turn these into a nice one-liner on the command line too:
perl -ne 'print if s/.*?;\s*//;'